1. ## Limit Question

Problem:

Lim [ cos(π+Δx) +1 ] / Δx
Δx -> 0

Hint: cos(x+y)=cosx cosy-sinx siny

My attempt:

[cosπ cosΔx + sinπ sinΔx + 1]/ Δx

[-1 cosΔx + 0 sinΔx +1 ]/ Δx

(-cosΔx + 1)/ Δx

I don't know what to do from here. I want to get rid of Δx from the denominator, but I don't know how to. Can someone please help me?

2. Use l'Hopitals?

$\displaystyle \lim\limits_{x\to0} \frac{-cos(x)+1}{x}=\lim\limits_{x\to0} \frac{sin(x)}{1} = 0$

Is that what you mean?

3. I wasn't taught that rule yet, so I doubt that is the way she wanted us to solve the problem. Can you please explain what you did to get that answer though?

4. Originally Posted by Power of One
I wasn't taught that rule yet, so I doubt that is the way she wanted us to solve the problem. Can you please explain what you did to get that answer though?
L'Hospital's Rule states that if a limit is of indeterminate form $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$, then

$\displaystyle \lim_{x \to 0}\frac{f(x)}{g(x)} = \lim_{x \to 0}\frac{f'(x)}{g'(x)}$.

$\displaystyle \lim_{\Delta x \to 0}\frac{-\cos{\Delta x} + 1}{\Delta x} = \lim_{\Delta x \to 0}\frac{\frac{d}{d\Delta x}(- \cos{\Delta x} + 1)}{\frac{d}{d\Delta x}(\Delta x)}$

$\displaystyle = \lim_{\Delta x \to 0}\frac{\sin{\Delta x}}{1}$

$\displaystyle = 0$

As it is, you don't need to resort to L'Hospital's Rule, as the original limit can also be solved using the sandwich theorem. Use Google to find some more info.

5. sin(x)/1 goes to 0 as x goes to 0.

6. So it does. I had the famous limit $\displaystyle \lim_{x \to 0}\frac{\sin{x}}{x}$ in my head at the time...